How much heat is required to convert 10g of ice at -20°C into steam at 100°C is(specific heat of ice=0.5cal/g-°C?

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14
Apr 12, 2017

Answer:

#7300# calories

Explanation:

When #10g# of ice at #-20°C# s being converted into steam at #100°C#, there are four stages.

  1. Ice at #-20°C# to ice at #0°C# - here it continues to be in the same state i.e. ice and hence heat required is #"mass"xx"specific heat"xx"change in temperature"# Specific heat for ice is #0.5# cal/g-°C.
  2. Then from ice at #0°C# to water at #0°C# and heat required is #"mass"xx"latent heat"#, this latent heat required is for conversion of each unit mass of substance. From ice to water it is #80"cal per gram"#
  3. Water at #0°C# to water at #100°C# - here it continues to be in the same state i.e. water and hence heat required is #"mass"xx"specific heat"xx"change in temperature"# Specific heat for water is #1# cal/g-°C.
  4. Then from water at #100°C# to steam at #100°C# and heat required is #"mass"xx"latent heat"#, this latent heat required is for conversion of each unit mass of substance. From water to steam, it is #540"cal per gram"#

Hence, to find heat is required to convert #10g# of ice at #-20°C# into steam at #100°C#,

first calculate heat required to convert #10g# of ice at #-20°C# to ice at #0°C#. As specific heat of ice is #0.5"cal/g°C"#, this is

#10xx20xx0.5=100# cal.

heat required to convert #10g# of ice to #10g# of water at #0°C# is

#10xx80=800# cal - as latent heat is #80"cal/g"#

heat required to convert #10g# of water at #0°C# to #10g# of water at #100°C# is

#10xx100x1=1000# cal

heat required to convert #10g# of water at #1000°C# to #10g# of steam at #100°C# is

#10xx540=5400# cal - as latent heat is #540"cal/g"#

Hence, total heat required is #100+800+1000+5400=7300# calories

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